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if x^3+px+r and 3x^2+p have a common factor, then p^3/27+r^2/4=0

if x^3+px+r and 3x^2+p have a common factor, then p^3/27+r^2/4=0
 

Grade:9

3 Answers

Vikas TU
14149 Points
7 years ago
let x= a be the common factor of both the given eqns.
then x = a must satisfy both.
.i.e
a^3 +ap + r = 0 …..............(1)
and  3a^2 + p = 0
a = + root(-p/3)  and a = -root(-p/3)...........(2)   ====> p
substitute eqn. (2) in eqn. (1)
(-p/3)^(3/2) + p*(root(-p/3)) + r  =0 
(-p/3)^(3/2) + p*(root(-p/3))  = -r
squaring both sides and after divide the whole eqn. by 4 u will get the result. 
mycroft holmes
272 Points
7 years ago
Note that 3x2+p is the derivative of x3+px+r and hence the common root ‘a’ is at least a double root of the cubic. If a is a triple root then since sum of roots is 0, a=0=p=r and the statement is true.
 
If a is a double root, then the third root is -2a.
Hence product of roots = -r = -2a3…..........................1
Also, from the quadratic we have a2 = -p/3..................2 
From the above two eqns the given relation easily follows.
Agni
11 Points
5 years ago
Let x-a is the common factor.
Then putting X=a we get a^2= -p/3 from 2nd equn
Then 
Putting X=a in 1st eqn we get
a^3+pa+r=0
a(a^2+p)+r=0
a(-p/3+p)+r=0
a= - 3r/2p
 
Now put this value of a to the 1st eqn.
r^2/4 + p^3/27=0

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